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Jlenok [28]
4 years ago
7

Aqueous potassium phosphate was mixed with aqueous magnesium chloride, and a crystallized magnesium phosphate product was formed

. Consider the other product and its phase, and then write the balanced molecular equation for this precipitation reaction. Express your answer as a chemical equation including phases.
Chemistry
1 answer:
scoray [572]4 years ago
4 0

<u>Answer:</u> The other product formed is potassium chloride.

<u>Explanation:</u>

Precipitation reaction is defined as the chemical reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

The chemical equation for the reaction of potassium phosphate and magnesium chloride follows:

2K_3PO_4(aq.)+3MgCl_2(aq.)\rightarrow Mg_3(PO_4)_2(s)+6KCl(aq.)

By Stoichiometry of the reaction:

2 moles of aqueous solution of potassium phosphate reacts with 3 moles of aqueous solution of magnesium chloride to produce 1 mole of solid magnesium phosphate and 6 moles of aqueous solution of potassium chloride.

Hence, the other product formed is potassium chloride.

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What is the approximate volume of 280 g of chlorine gas (Cl2) at STP
Anettt [7]
<h3>Answer:</h3>

78 L Cl₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
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<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

280 g Cl₂

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.90 g/mol

[STP] 1 mol = 22.4 L

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                   \displaystyle 280 \ g \ Cl_2(\frac{1 \ mol \ Cl_2}{79.90 \ g \ Cl_2})(\frac{22.4 \ L \ Cl_2}{1 \ mol \ Cl_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                       \displaystyle 78.4981 \ L \ Cl_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

78.4981 L Cl₂ ≈ 78 L Cl₂

5 0
3 years ago
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3 0
4 years ago
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Arte-miy333 [17]
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7 0
3 years ago
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GaryK [48]

Answer:

Explanation:

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6 0
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What is the concentration of each ion in a solution that is prepared by dissolving 5.00 g of ammonium chloride in enough water t
const2013 [10]

Answer:

C = 0.08M

Explanation:

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